Oxidize ethanol and you get acetaldehyde (ethanal). Yet you cannot point to an electron that leaves the molecule. What changes is what one carbon is bonded to.
General chemistry defines oxidation as a loss of electrons and reduction as a gain. That definition is correct, but it is hard to apply to a reaction like this one. So organic chemists track bonds instead, which gives the same answers with far less arithmetic.
The working test
Oxidize propan-2-ol and you get propanone (acetone). Watch the middle carbon. It loses its C–H bond, and a second bond to oxygen takes its place, making the C=O.
Now run the change backward, from propanone to propan-2-ol. The middle carbon gains a C–H bond and loses one of its bonds to oxygen. That is a reduction.
On this page an electronegative atom means oxygen, nitrogen or a halogen (F, Cl, Br or I). Each one pulls bonding electrons more strongly than carbon does, as Electronegativity showed. The two directions give the working test:
Reduction = the reverse: losing bonds to oxygen, or gaining bonds to hydrogen.
Adding to an alkene
Add HBr to propene and you get 2-bromopropane. One alkene carbon gains a C–H bond, which counts toward reduction. The other gains a C–Br bond, which counts toward oxidation. Count over both carbons together and the two changes cancel, so the reaction is neither an oxidation nor a reduction.
Bromine behaves differently. Br2 puts a bromine on each alkene carbon and a hydrogen on neither, so both carbons move toward oxidation. Adding Br2 to an alkene is an oxidation.
Hydrogenation is the mirror image. H2 puts a hydrogen on each carbon and adds no bond to oxygen, so hydrogenation is a reduction. Its name does not say so, which is why it is easy to miss.
The same count sorts every alkene addition from Alkenes & Alkynes. An addition that delivers one hydrogen and one electronegative atom is neither. One that puts an electronegative atom on both carbons is an oxidation.
| Addition | What the two alkene carbons gain | Verdict |
|---|---|---|
| Hydrohalogenation (HBr, HCl) | H on one, a halogen on the other | neither |
| Acid-catalyzed hydration | H on one, OH on the other | neither |
| Oxymercuration–demercuration | H on one, OH on the other | neither |
| Hydroboration–oxidation | H on one, OH on the other | neither |
| Halogenation (Br2, Cl2) | a halogen on both | oxidation |
| Halohydrin formation (Br2 in water) | OH on one, Br on the other | oxidation |
| Epoxidation (mCPBA) | a bond to the same O on both | oxidation |
| Dihydroxylation (OsO4) | OH on both | oxidation |
| Hydrogenation (H2, Pd) | H on both | reduction |
Hydroboration–oxidation has “oxidation” in its name, yet the whole sequence is neither. The name describes its second step, which turns the C–B bond into a C–O bond. Counted from the alkene to the alcohol, one carbon gains H and the other gains OH.
Putting a number on a carbon
Sometimes you want a number, not just a direction. Take methanol, CH3OH. Its carbon has three bonds to hydrogen and one to oxygen. Score each bond to hydrogen −1 and the bond to oxygen +1, and the carbon’s oxidation state is −3 + 1 = −2. The general rule:
- Each bond to an electronegative atom (O, N or a halogen) counts +1.
- Each bond to hydrogen counts −1.
- Each bond to another carbon counts 0, so atoms further along the chain do not affect the score.
The scores come from electronegativity. Hand each bond’s two electrons to the more electronegative of its two atoms. Oxygen takes both electrons of a C–O bond, which leaves carbon one electron short: +1. Carbon is slightly more electronegative than hydrogen, so it takes both electrons of a C–H bond and ends up one electron over: −1. Two carbons share a C–C bond evenly: 0. The total is the carbon’s oxidation state in the general-chemistry sense, worked out one carbon at a time. A bond to an atom less electronegative than carbon, such as boron or a metal, scores −1 for the same reason as a bond to hydrogen.
A double bond is two bonds, and each one is scored. The C=O of a carbonyl group scores +2, and the C≡N of a nitrile scores +3.
Find the oxidation state of the CH2 carbon of ethanol, CH3CH2OH. Then find it for the same carbon in acetaldehyde, CH3CHO, which is what ethanol gives when it is oxidized.
Step 1 — list the four bonds in ethanol. That carbon has one bond to the CH3 carbon, two to hydrogen and one to the oxygen of the OH.
Step 2 — score them. The C–C bond counts 0, and the two C–H bonds count −1 each. The C–O bond counts +1. The total is 0 − 2 + 1 = −1.
Step 3 — do the same for acetaldehyde. The carbon now has one bond to CH3 (0), one to H (−1) and a double bond to oxygen (+2). The total is 0 − 1 + 2 = +1.
Step 4 — read the change. The number rose from −1 to +1, by two, so this is a two-electron oxidation.
The carbon oxidation ladder
Score the one-carbon compounds from methane up to carbon dioxide and they fall on five levels:
| Compound | Carbon oxidation state | Class |
|---|---|---|
| CH₄ | −4 | alkane |
| CH₃OH | −2 | alcohol |
| CH₂O | 0 | aldehyde |
| HCO₂H | +2 | carboxylic acid |
| CO₂ | +4 | fully oxidized |
Each move up the table replaces one C–H bond with one more bond to oxygen, so the number rises by two. By the electron count, the carbon gives up two electrons each time, so every move up is a two-electron oxidation. Call these levels rungs, and the whole sequence the oxidation ladder: alkane, then alcohol, then aldehyde or ketone, then carboxylic acid, then CO2.
A rung is set by how many bonds the carbon has to electronegative atoms: none for an alkane, one for an alcohol, two for an aldehyde or ketone, three for a carboxylic acid and four for CO2. A C=O counts as two of these bonds, which is why an aldehyde sits a whole rung above an alcohol. Functional group priority used this same count to rank the suffixes and called it the carbon’s oxidation level; this page calls each level a rung.
Every compound with the same count shares a rung. An alcohol and an alkyl halide each have one bond from carbon to an electronegative atom, so they sit on the same rung. Turning an alcohol into an alkyl bromide with PBr3, as in Alcohol reactions, is therefore a substitution, not a redox reaction.
A carboxylic acid, an ester, an amide, an acid chloride and a nitrile each have three such bonds, so they share a rung too. Turning one of them into another needs no oxidant and no reductant; the reactions that do it, chiefly nucleophilic acyl substitution, are the subject of Carboxylic Acids & Derivatives. Getting off that rung, down to an aldehyde or an alcohol, does need a reducing agent.
The number on a rung depends on what else the carbon holds. The table uses one-carbon compounds, so formic acid, HCO2H, reads +2: its carbon still carries a hydrogen. In acetic acid, CH3CO2H, a carbon sits where that hydrogen was, and a C–C bond scores 0 instead of −1, so the acid carbon reads +3. Methyl acetate, acetamide and acetonitrile read +3 for the same reason. The same thing makes ethanol’s carbon −1 where methanol’s is −2. The rung stays the same, and so does the rise of two from one rung to the next.
The ladder predicts how far an alcohol can climb
Oxidize propan-1-ol and it climbs two rungs: first to propanal, then to propanoic acid. Propan-2-ol climbs one rung, to propanone. 2-Methylpropan-2-ol is not oxidized at all.
The difference lies in the carbon that carries the OH, called the carbinol carbon. Each oxidation removes one hydrogen from that carbon and puts a bond to oxygen in its place. So the number of hydrogens on the carbinol carbon is the number of rungs the alcohol can climb.
As in Functional groups, alcohols are sorted by this same carbon. In a primary (1°) alcohol the carbinol carbon is bonded to one other carbon, so it carries two hydrogens and can be oxidized twice, to an aldehyde and then to a carboxylic acid. In a secondary (2°) alcohol it is bonded to two carbons and carries one hydrogen, so it stops at a ketone. In a tertiary (3°) alcohol it is bonded to three carbons and has no hydrogen left, so an ordinary oxidant leaves it alone.
Recognizing an oxidant or a reductant on sight
Stir propan-2-ol with Jones reagent, CrO3 in aqueous sulfuric acid, and you get propanone. The alcohol’s carbon rises by two. Meanwhile the orange chromium(VI) of the reagent becomes green chromium(III), a fall of three. Every oxidation is paired with a reduction like this: when the carbon goes up, something in the reagent comes down.
So an oxidant contains an atom that can drop to a lower oxidation state, and a reductant has electrons or a hydride to give. The table lists the ones this course uses; KMnO4, OsO4, O3 and mCPBA first appeared in Alkenes & Alkynes.
| Reagent | Role | What happens to the reagent |
|---|---|---|
| Jones reagent (CrO3, H2SO4, water), or PCC (pyridinium chlorochromate, a Cr(VI) reagent used without water) | oxidant | Cr(VI) falls to Cr(III); Jones reagent turns from orange to green |
| KMnO4 | oxidant | Mn(VII) falls to Mn(IV), as brown MnO2 in neutral or basic water |
| OsO4 | oxidant | Os(VIII) falls to Os(VI) |
| O3 and mCPBA | oxidant | each hands an oxygen atom to the substrate |
| NaBH4 and LiAlH4 | reductant | each gives a hydride ion, H− (a hydrogen with two electrons) |
| H2 with Pd, Pt or Ni | reductant | the metal surface delivers the two hydrogen atoms of H2 |
| Na in liquid NH3 | reductant | sodium gives electrons directly and becomes Na+ |
| Zinc amalgam, Zn(Hg), in HCl | reductant | zinc gives electrons and becomes Zn2+ (the Clemmensen reduction, in Reducing carbonyls) |
Most of the oxidants are a metal in a high oxidation state, or a reagent carrying an oxygen atom it can hand over. The reductants are the mirror image: a source of H−, H2 on a metal surface, or a metal that gives up electrons.
That pattern is a fast check for a reagent you do not recognize. Look for an atom in an unusually high oxidation state, or a spare oxygen: that reagent is an oxidant. Look for a source of H−, H2 or electrons: that one is a reductant.
The color change of chromium made these reactions useful as tests long before anyone drew a mechanism. The classic breathalyzer oxidized the ethanol in a breath sample with orange Cr(VI) and measured the green Cr(III) that formed.
What carries forward
The ladder tells you which way a change goes; the reagent decides where it stops. Propan-1-ol gives propanal with PCC but propanoic acid with Jones reagent. Adding more PCC or waiting longer does not change that; Oxidizing alcohols explains why. Going down the ladder works the same way. NaBH4 reduces a ketone and leaves an ester alone, while LiAlH4 reduces both, as Reducing carbonyls covers in full. So for every reagent in this chapter, ask two questions: where does it stop, and what else in the molecule does it attack?
The ladder also shapes synthesis planning. When a change spans two rungs and no single reagent makes it, the route takes two steps.